mAth probLem soLving ,pLease heLp me !!!?

Question

A boAt left the dock for Manitou Island at 14:00 travelling at km/h .An hour later ,another boat left the same dock.also for Manitou Island ,travelling at 30 km/h. How long did it take the faster boat to pass the slower one.?

A boAt left the dock for Manitou Island at 14:00 travelling at 20 km/h .An hour later ,another boat left the same dock.also for Manitou Island ,travelling at 30 km/h. How long did it take the faster boat to pass the slower one.?

Answer 1

You haven't stated what the speed of the first boat is.

Let's say it's "a" km/h. Let's also say the time it takes for the second boat to catch the first is "t". Then the distance of the first boat from the dock at that time will be. at + a (It has already gone for 1 hour. That's where the "+a" comes in). The second boat will then be at position 30t. These two distances have to be equal. So,

at+a = 30t

(30-a)t = a
t = a/(30-a)

You'll notice that there is no positive solution for t when a is greater than or equal to 30. In other words, if the speed of the first boat is greater than or equal to the speed of the second boat, the second boat will never catch it.

Answer 2

let t be the time the first boat left, a hour later or t - 1, the second boat left.

When the second boat catches the first boat, their distances will be the same

x = vt

x = 20t
x = 30(t-1)


20t = 30(t-1)
20t = 30t - 30
-10t = -30
t = 3hr

starting from the time of the second boat, it takes the second boat 2 hours to catch the first boat

Answer 3

Since you do not say the speed of the boat leaving first, we can only assume it is z km/h.

So z +zt = 30 t
30t -zt = z
t(30-z)=z
t = z/(30-z)

So just plug the missing speed of the 1st boat into above equation for z and solve for t.

Answer 4

missing a bit of information!!